Chapter 23, Problem 23.107P

(a)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Iron (III) ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation of Solution

Given,

Iron (III) ion reacts with hypochlorite ion in basic solution to form ferrate ion

    $\left(2Fe{O}_4^{2-}\right),C{l}^{-},and{\text{ H}}_{2}O.$

The balanced equation is

$2F{e}^{3+}{}_{(aq)}+6OC{l}_{(aq)}^{-}+4O{H}^{-}{}_{(aq)}\to 2Fe{O}_{4(aq)}^{2-}+6C{l}_{(aq)}^{-}+2H_2{O}_{(l)}$

Based on the chemical equation, the total number of iron oxidation is +6 each Iron has +3 oxidation state.

(b)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Manganese and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation of Solution

Given,

Potassium hexacyanomanganate (II) reacts with $K$ metal to form $K_6\left[Mn{\left(CN\right)}_6\right].$

The balanced equation is

$K_4{\left[Mn{\left(CN\right)}_6\right]}_{(s)}+2K_{(s)}\to K_6{\left[Mn{\left(CN\right)}_6\right]}_{(s)}$

Based on the chemical equation six potassium cation balanced by six cyanide ions.  Hence, the oxidation of Manganese is zero.

(c)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Cobalt ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation of Solution

Given,

Heating sodium superoxide $\left(Na{O}_2\right)withC{o}_3{O}_{4}producesN{a}_{4}Co{O}_{4}and{O}_{2}gas.$

The balanced equation is

  $12Na{O}_2{}_{(s)}+C{o}_3{O}_4{}_{(s)}^{}\to 3N{a}_{4}Co{O}_{4(s)}+8O_2{}_{(g)}$

Based on the chemical equation oxygen eight negative charge balance by four sodium and another four balance by Cobalt.  Therefor the Cobalt oxidation state is four.

(d)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Vanadium ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation of Solution

Vanadium (III) chloride reacts with $Na$ metal under a $CO$ atmosphere to produce

    $Na\left[V{\left(CO\right)}_6\right]andNaCl.$

The balanced equation is

$VC{l}_3{}_{(s)}+4N{a}_{(s)}^{}+6C{O}_{(g)}\to Na{[V{(CO)}_6]}_{(s)}^{}+3NaC{l}_{(s)}$

Based on the chemical equation sodium has one positive charge that positive charge balanced by vanadium so the vanadium has -1 charge.

Therefore, the oxidation state of Vanadium is -1

(e)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Nickel ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation of Solution

Given,

Barium peroxide reacts with nickel (II) ions in basic solution to produce $BaNi{O}_3.$

The balanced equation is

$2F{e}^{3+}{}_{(aq)}+6OC{l}_{(aq)}^{-}+4O{H}^{-}{}_{(aq)}\to 2Fe{O}_{4(aq)}^{2-}+6C{l}_{(aq)}^{-}+2H_2{O}_{(l)}$

  $\begin{array}{l}BaNi{O}_3.\\ +2+x+3(-2)=0\\ +2+x-6=0\\ x-4=0\\ x=+4\end{array}$

Hence, the oxidation of Nickel is +4

(f)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Cobalt ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation of Solution

Bubbling $CO$ through a basic solution of cobalt (II) ion Produces ${\left[Co{\left(CO\right)}_4\right]}^{-},C{O}_3^{2-},$ and water.

The balanced equation is

$11C{O}_{(g)}+12O{H}_{(aq)}^{-}+2C{o}_{(aq)}^{2+}\to 2{[Co{(CO)}_4]}_{(aq)}^{-}+3C{O}_{3(aq)}^{2-}+6H_2{O}_{(l)}$

${[Co{(CO)}_4]}_{(aq)}^{-}$ here the ligand is neutral.  So the oxidation state of the Cobalt is -1.

(g)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Copper ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation of Solution

Heating cesium tetrafluorocuprate (II) with $F_2$ gas under pressure gives $C{s}_{2}Cu{F}_6.$

The balanced equation is

$C{s}_2{[Cu{F}_4]}_{(s)}+F_{2(g)}\to C{s}_{2}Cu{F}_6{}_{(s)}$

  $\begin{array}{l}C{s}_{2}Cu{F}_6\\ 2(+1)+x+6(-1)=0\\ +2+x-6=0\\ x-4=0\\ x=+4\end{array}$

Hence, the oxidation state of the copper is +4.

(h)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Tantalum ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation of Solution

Given,

Heating tantalum (V) chloride with $Na$ metal produces $NaClandT{a}_{6}C{l}_{15},$ in which half of the $Ta$ is in the +2 state.

The balanced equation is

$6TaC{l}_5{}_{(s)}+15N{a}_{(s)}\to 15NaC{l}_{(s)}+T{a}_{6}C{l}_{15}{}_{(s)}$

$T{a}_{6}C{l}_{15}{}_{(s)}$ here the ligand is chloride, half of the Tandalum +2 oxidation state and another half of the tantalum is +3 oxidation state.

Hence, 3 Tantalum is +3 oxidation state and 3 Tantalum is +2 oxidation state.

(i)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Nickel ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

Explanation of Solution

Potassium tetracyanonickelate (II) reacts with hydrazine $\left(N_2{H}_4\right)$ in basic solution to form

    $K_4\left[N{i}_2{\left(CN\right)}_6\right]and{N}_{2}gas.$

The balanced equation is

$4K_2{[Ni{(CN)}_4]}_{(s)}+N_2{H}_{4(aq)}+4O{H}_{(aq)}^{-}\to 2K_4{[N{i}_2{(CN)}_6]}_{(s)}+N_{2(g)}+4H_2{O}_{(l)}+4C{N}_{(aq)}^{-}$

$\begin{array}{l}{K}_4{[}_N\\ 4(+1)+2x+6(-1)=0\\ 4+2x-6=0\\ 2x-2=0\\ 2x=+2\\ x=+1\end{array}$

Hence, the oxidation state of the Nickel is +1.